About questions 1 to 3, you just have to equals the argument of the time-scaled/shifted function to each point of the original function. That is:
1. t1/10=-1 ==> t1=-10, t2/10=1 ==> t2=10, t3/10=2 ==> t3=20 you graph is correct (pure time scaling)
2. t1/10-5=-1 ==> t1=40, t2/10-5=1 ==> t2=60, t3/10-5=2 ==> t3=70 that is shift right by 5 before, then scale by 10
3. (t1-5)/10=-1 ==> t1=-5, (t2-5)/10=1 ==> t2=15, (t3-5)/10=2 ==> t3=25 that is scale by 10 before, then shift right by 5
so you inverted graph 2 with graph 3.
About part 4 I think is asked to find which one of the three functions is periodic.
Our function has a total duration of 2-(-1)=3
the first is the summation of infinite x(t) shifted by 3 so is exactly x(t) periodicized (I don't know if this term is correct in English)
the second one is the summation of infinite x(t) shifted by 4, the is a periodic function different from x(t) there is an added
the third one is the summation of infinite x(t) shifted by 2, then there will be superposition between a shifted x(t) and the following